Question

The number of atoms present in one mole of an element is equal to Avogadro number. Which of the following element contains the greatest number of atoms?

• A. 4g He
• B. 0.40g Ca
• C. 12g He
• D. 46g Na

Answer 1

The correct option is C 12g He

Calculating the number of moles of elements


No. of moles
$=\frac{\text{Given mass of a substance}}{\text{Molar mass of a substance}}$

A. 4 g He (Molar mass = $4gmo{l}^{-1})$
No. of moles $=\frac{4}{4}=1mol$

B. 46 g Na (Molar mass = $23gmo{l}^{-1})$
No. of moles $=\frac{46}{23}=2mol$

C. 0.40 g Ca (Molar mass = $40gmo{l}^{-1})$
No. of moles $=\frac{0.40}{40}=0.01mol$

D. 12 g He (Molar mass = $4gmo{l}^{-1})$
No. of moles $=\frac{12}{4}=3mol$

Calculating number of atoms of elements

Number of atoms present in one mole of element is equal to Avogadro's number.

Number of atoms in 1 mole
$=6.022\times {10}^{23}atoms$

Number of atoms in an element
= no. of moles $\times 6.022\times {10}^{23}$

A. No. of moles of He = 1 mol
No. of atoms of He $=1\times 6.022\times {10}^{23}$
$=6.022\times {10}^{23}$ atoms

B. No. of moles of Na = 2 mol
No. of atoms of Na $=2\times 6.022\times {10}^{23}$
$=12.044\times {10}^{23}$ atoms

C. No. of moles of Ca = 0.01 mol
No. of atoms of Ca $=0.01\times 6.022\times {10}^{23}$
$=6.022\times {10}^{23}$ atoms

D. No. of moles of He = 3 mol
No. of atoms of He $=3\times 6.022\times {10}^{23}$
$=18.066\times {10}^{23}$ atoms